Integration of an inverse sqrt composite function

1. The problem statement, all variables and given/known data“Geologist A” at the bottom of a cave signals to his colleague “Geologist B” at the surface by pushing a 11.0 kg box of samples from side to side. This causes a transverse wave to propagate up the 77.0 m rope. The total mass of the rope is 14.0 kg. Take g = 9.8 m/s².

How long does it take for the wave to travel from the bottom of the cave to the surface?[Hint: Find an analytic expression v(z) for the wave speed as a function of distance. Then use the fact that at any given point on the rope the time dt taken to travel a small distance dz is given by: dt=dz/v(z). Then integrate to obtain the total travel time. ]

'u' in your solution is supposed to be the mass density of the rope. It's not mR/z. The rope doesn't have a variable density, it's mR/(total length of rope), a constant. Nice problem presentation, by the way.

Hi Dick, z is the length of the rope (77.0m as in the problem) it was just the letter they used in the formula sheet so I carried it forth.

How was my integration? I don't think I know how to integrate nested functions (I assume it's something like the reverse of the chain-rule?).

z in your problem is the variable indicating length along the rope. I mean that u=14 kg/(77 m). It doesn't have the variable of integration in it. Until that gets fixed there isn't any point in discussing the integral.

z in your problem is the variable indicating length along the rope. I mean that u=14 kg/(77 m). It doesn't have the variable of integration in it. Until that gets fixed there isn't any point in discussing the integral.

Not too good. You're coming up with some pretty bizarre integration rules which aren't in the book. You have to do the change of variable thing. Eg to integrate 1/(a+bz)^(1/2) I would say v=(a+bz), so dv=b*dz. This turns the integral into 1/v^(1/2)*dv*(1/b). Now it's just integrating v^(-1/2). Does that sound familiar?

Not too good. You're coming up with some pretty bizarre integration rules which aren't in the book. You have to do the change of variable thing. Eg to integrate 1/(a+bz)^(1/2) I would say v=(a+bz), so dv=b*dz. This turns the integral into 1/v^(1/2)*dv*(1/b). Now it's just integrating v^(-1/2). Does that sound familiar?

Unfortunately not very familiar at all. I haven't really dealt with composite integrals. But I'll try:

a = g.z
b = (mB.g)/u
1/(a+bz)^(1/2)

v=(a+bz), so dv=b*dz.
This turns the integral into 1/v^(1/2)*dv*(1/b).
Integrating v^(-1/2).

Take a break and clear your head. While your at it look back at integration by substitution in a calc text. I can roughly see what you are trying to do - but you still seem to be trying to do some kind of a chain rule. And the (1/b) factor in the example becomes (1/g) in the problem, right? Do you see where it's coming from? And after the integration is done and the dz is gone you should also drop the integral sign - it looks pretty confusing otherwise.